Answer:
Step-by-step explanation:
Hello!
Since you didn't upload the data, I'll use my own data to solve the example. The steps will be the same, only the results will change. (see attachment) The hypothesis test will be made using a 5% significance level.
The objective of this exercise is to test if there is an association between two variables:
X₁: The subject is vaccinated, categorized "Yes" and "No"
X₂: The subject got the disease, categorized "Yes" and "No"
These two variables of interest are qualitative categorical and each one of them has two categories.
To test if the variables are associated, considering the type of variables they are, you have to apply a Chi Square test of Independence, where the statistic hypotheses are:
H₀: [tex]P_{ij}= P_{i.} * P_{.j}[/tex] ∀ i= 1, 2 and j= 1, 2
H₁: The variables, vaccination and disease status, are not independent.
α: 0.05
The statistic is
X²= [tex]{r} \atop {i=1} \right.[/tex]∑[tex]{c} \atop {j=1} \right.[/tex]∑[tex]\frac{(O_{ij}-E_{ij})^2}{E_{ij}}[/tex]≈[tex]X^2_{(r-1)(c-1)}[/tex]
i= values in rows
j= values in columns
r= total number of rows
c= total number of columns
This type of test is always one-tailed to the right, meaning that you will reject the null hypothesis to high values of Chi Square. There is only one critical value:
[tex]X^2_{(r-1)(c-1);1-\alpha }= X^2_{(2-1)(2-1);1-0.05}= X^2_{1*1;0.95}= 3.841[/tex]
The decision rule will be:
If [tex]X^2_{H_0}[/tex] ≥ 3.841, reject the null hypothesis.
If [tex]X^2_{H_0}[/tex] < 3.841, do not reject the null hypothesis.
Using the p-value approach, the decision rule is always the same:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
Before calculating [tex]X^2_{H_0}[/tex], you have to calculate the expected frequencies for all categories using the formula:
Ê[tex]_{ij}[/tex]= [tex]\frac{O_{i.}*O_{.j}}{n}[/tex]
Ê₁₁= [tex]\frac{O_{1.}*O_{.1}}{n}= \frac{82*100}{200} = 41[/tex]
Ê₁₂= [tex]\frac{O_{1.}*O_{.2}}{n}= \frac{82*100}{200} = 41[/tex]
Ê₂₁= [tex]\frac{O_{2.}*O_{.1}}{n}= \frac{118*100}{200} = 59[/tex]
Ê₂₂= [tex]\frac{O_{2.}*O_{.2}}{n}= \frac{118*100}{200} = 59[/tex]
[tex]X^2_{H_0}= \frac{(42-41)^2}{41} + \frac{(40-41)^2}{41} + \frac{(58-59)^2}{59} + \frac{(60-59)^2}{59}= 0.0827[/tex]
The p-value for this test is the probability of obtaining a value as extreme as [tex]X^2_{H_0}[/tex]= 0.0827:
P(X₁² ≥ 0.0827)= 1 - P(X₁² < 0.0827)= 1 - 0.2264= 0.7736
Ata 5% significance level, the decision is to not reject the null hypothesis. You can conclude that the vaccination and the disease status of the subjects are not related. The new vaccine does not affect the chances of the subjects getting the disease.
I hope this helps!
